Equivalent Length vs. Minor Pressure Head Loss in Pipe and Duct Components
Minor pressure and head loss in pipes vs. equivalent length in tubes and duct systems.
The total energy per mass unit in a given point in a fluid flow consists of elevation (potential) energy , velocity (kinetic) energy and pressure energy .
The Energy Equation states that energy can not disappear  the energy upstream in the fluid flow will always be equal to the energy downstream in the flow and the energy loss between the two points.
E_{1} = E_{2}+ E _{ loss } (1)
where
E_{1} = energy upstream (J/kg, Btu/lb)
E_{2}= energy downstream (J/kg, Btu/lb)
E _{ loss } = energy loss (J/kg, Btu/lb)
The energy in a specific point in the flow
E _{ flow } = E _{ pressure } + E _{ kinetic } + E _{ potential } (2)
where
E _{ pressure } = p / ρ = pressure energy (J/kg, Btu/lb)
E _{ kinetic } = v^{2}/ 2 = velocity (kinetic) energy (J/kg, Btu/lb)
E _{ potential } = g h = elevation (potential) energy (J/kg, Btu/lb)
E _{ loss } = Δ p _{ loss } / ρ = major and minor energy loss in the fluid flow (J/kg, Btu/lb)
p = pressure in fluid (Pa (N/m^{2}), psi (lb/in^{2}))
Δ p _{ loss } = major and minor pressure loss in the fluid flow (Pa (N/m^{2}), psi (lb/in^{2}))
ρ = density of fluid (kg/m^{3}, slugs/ft^{3} )
v = flow velocity (m/s, ft/s)
g = acceleration of gravity (m/s^{2}, ft/s^{2})
h = elevation (m, ft)
Eq. 1 and 2 can be combined to express the equal energies in two different points in a stream line as
p_{1} / ρ + v_{1}^{2}/ 2 + g h_{1} = p_{2}/ ρ + v_{2}^{ 2 } / 2 + g h_{2}+ Δp _{ loss } / ρ (3)
or alternatively
p_{1} + ρ v_{1}^{2}/ 2 + ρ g h_{1} = p_{2}+ ρ v_{2}^{ 2 } / 2 + ρ g h_{2}+ Δp _{ loss } (3b)
For a horizontal steady state flow v_{1} = v_{2} and h_{1} = h_{2},  and (3b) can be simplified to:
Δ p _{ loss } = p_{1}  p_{2}(3c)
The pressure loss is divided in
 major loss due to friction and
 minor loss due to change of velocity in bends, valves and similar
The major friction loss in a pipe or tube depends on the flow velocity, pipe or duct length, pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct, and whether the flow us turbulent or laminar  the Reynolds Number of the flow. The pressure loss in a tube or duct due to friction, major loss, can be expressed as:
Δ p _{ major_loss } = λ (l / d _{ h } ) (ρ v^{2}/ 2) (4)
where
Δ p _{ major_loss } = major friction pressure loss (Pa, (N/m^{2}), lb/ft^{2})
λ = friction coefficient
l = length of duct or pipe (m, ft)
d _{ h } = hydraulic diameter (m, ft)
Eq. (3) is also called the D'ArcyWeisbach Equation . (3) is valid for fully developed, steady, incompressible flow .
The minor or dynamic loss depends flow velocity, density and a coefficient for the actual component.
Δ p _{ minor_loss } = ξ ρ v^{2} / 2 (5)
where
Δ p _{ minor_loss } = minor pressure loss (Pa (N/m^{2}), lb/ft^{2})
Head and Head Loss
The Energy equation can be expressed in terms of head and head loss by dividing each term by the specific weight of the fluid. The total head in a fluid flow in a tube or a duct can be expressed as the sum of elevation head , velocity head and pressure head .
Note ! The heads in the equations below are based on the fluid itself as a reference fluid. Read more about head here .
p_{1} / γ + v_{1}^{2}/ 2 g + h_{1} = p_{2}/ γ + v_{2}^{ 2 } / 2 g + h_{2}+ Δh _{ loss } (6)
where
Δ h _{ loss } = head loss (m "fluid", ft "fluid")
γ = ρ g = specific weight of fluid (N/m^{3}, lb/ft^{3} )
For horizontal steady state flow v_{1} = v_{2} and p _{1} = p_{2},  (4) can be transformed to:
h _{ loss } = h_{1}  h_{2}(6a)
where
Δ h = p / γ = head (m "fluid", ft "fluid")
The major friction head loss in a tube or duct due to friction can be expressed as:
Δ h _{ major_loss } = λ (l / d _{ h } ) (v^{2}/ 2 g) (7)
where
Δ h _{ loss } = head loss (m, ft)
The minor or dynamic head loss depends flow velocity, density and a coefficient for the actual component.
Δ p _{ minor_loss } = ξ v^{2}/ (2 g) (8)
Friction Coefficient  λ
The friction coefficient depends on the flow  if it is
 laminar,
 transient or
 turbulent
and the roughness of the tube or duct.
To determine the friction coefficient we first have to determine if the flow is laminar, transient or turbulent  then use the proper formula or diagram.
Friction Coefficient for Laminar Flow
For fully developed laminar flow the roughness of the duct or pipe can be neglected. The friction coefficient depends only the Reynolds Number  Re  and can be expressed as:
λ= 64 / Re (9)
where
The flow is
 laminar when Re < 2300
 transient when 2300 < Re < 4000
 turbulent when Re > 4000
Friction Coefficient for Transient Flow
If the flow is transient  2300 < Re < 4000  the flow varies between laminar and turbulent flow and the friction coefficient is not possible to determine.
Friction Coefficient for Turbulent Flow
For turbulent flow the friction coefficient depends on the Reynolds Number and the roughness of the duct or pipe wall. On functional form this can be expressed as:
λ = f( Re, k / d _{ h } ) (10)
where
k = absolute roughness of tube or duct wall (mm, ft)
k / d _{ h } = the relative roughness  or roughness ratio
Roughness for materials are determined by experiments. Absolute roughness for some common materials are indicated in the table below
Surface  Absolute Roughness  k  

(10^{3} m)  (feet)  
Copper, Lead, Brass, Aluminum (new)  0.001  0.002  3.3  6.7 10^{6} 
PVC and Plastic Pipes  0.0015  0.007  0.5  2.33 10 ^{ 5 } 
Epoxy, Vinyl Ester and Isophthalic pipe  0.005  1.7 10 ^{ 5 } 
Stainless steel, bead blasted  0.001  0.006  (0.00328  0.0197) 10^{3} 
Stainless steel, turned  0.0004  0.006  (0.00131  0.0197) 10^{3} 
Stainless steel, electropolished  0.0001  0.0008  (0.000328  0.00262) 10^{3} 
Steel commercial pipe  0.045  0.09  1.5  3 10^{4} 
Stretched steel  0.015  5 10 ^{ 5 } 
Weld steel  0.045  1.5 10^{4} 
Galvanized steel  0.15  5 10^{4} 
Rusted steel (corrosion)  0.15  4  5  133 10^{4} 
New cast iron  0.25  0.8  8  27 10^{4} 
Worn cast iron  0.8  1.5  2.7  5 10^{3} 
Rusty cast iron  1.5  2.5  5  8.3 10^{3} 
Sheet or asphalted cast iron  0.01  0.015  3.33  5 10 ^{ 5 } 
Smoothed cement  0.3  1 10^{3} 
Ordinary concrete  0.3  1  1  3.33 10^{3} 
Coarse concrete  0.3  5  1  16.7 10^{3} 
Well planed wood  0.18  0.9  6  30 10^{4} 
Ordinary wood  5  16.7 10^{3} 
The friction coefficient  λ  can be calculated by the Colebrooke Equation :
1 / λ ^{ 1/2 } = 2,0 log _{ 10 } [ (2,51 / (Re λ ^{ 1/2 } )) + (k / d _{ h } ) / 3,72 ] (11)
Since the friction coefficient  λ  is on both sides of the equation, it must be solved by iteration. If we know the Reynolds number and the roughness  the friction coefficient  λ  in the particular flow can be calculated.
A graphical representation of the Colebrooke Equation is the Moody Diagram :
 The Moody Diagram  The Moody diagram in a printable format.
With the Moody diagram we can find the friction coefficient if we know the Reynolds Number  Re  and the
Relative Roughness Ratio  k / d _{ h }
In the diagram we can see how the friction coefficient depends on the Reynolds number for laminar flow  how the friction coefficient is undefined for transient flow  and how the friction coefficient depends on the roughness ratio for turbulent flow.
For hydraulic smooth pipes  the roughness ratio limits zero  and the friction coefficient depends more or less on the Reynolds number only.
For a fully developed turbulent flow the friction coefficient depends on the roughness ratio only.
Example  Pressure Loss in Air Ducts
Air at 0 ^{o}C is flows in a 10 m galvanized duct  315 mm diameter  with velocity 15 m/s .
Reynolds number can be calculated:
Re = d _{ h } v ρ / μ (12)
where
Re = Reynolds number
v = velocity (m/s)
ρ = density of air (kg/m^{3} )
μ = dynamic or absolute viscosity ( Ns/m^{2})
Reynolds number calculated:
Re = (15 m/s) (315 mm) (10^{3} m/mm ) (1.23 kg/m^{3} ) / (1.79 10 ^{ 5 } Ns/m^{2})
= 324679 (kgm/s^{2})/N
= 324679 ~ Turbulent flow
Turbulent flow indicates that Colebrooks equation (9) must be used to determine the friction coefficient  λ .
With roughness  ε  for galvanized steel 0.15 mm , the roughness ratio can be calculated:
Roughness Ratio = ε / d _{ h }
= (0.15 mm) / (315 mm)
= 4.76 10 ^{4}
Using the graphical representation of the Colebrooks equation  the Moody Diagram  the friction coefficient  λ  can be determined to:
λ = 0.017
The major loss for the 10 m duct can be calculated with the DarcyWeisbach Equation (3) or (6):
Δp _{ loss } = λ ( l / d _{ h } ) ( ρ v^{2}/ 2 )
= 0.017 ((10 m) / (0.315 m)) ( (1.23 kg/m^{3} ) (15 m/s)^{2}/ 2 )
= 74 Pa (N/m^{2})
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